[seqfan] Re: Prime partial sum of digits

[Chris Thompson]

On Nov 9 2015, Eric Angelini wrote:

>Hello Bob,
>
>> S = 2,1,4,6,40,20,46,8,4,24,60,...
>> P = 2 3 7 1 11 11 22 3 4 44 55
>>           3 77 99 39 7 1 37 33
>
>... a(n) is the nth term of S [I hesitated naming it S(n)]
>... P(4) is 13 (cumulative sum 2+1+4+6 = 13 --> to be read vertically under "6")
>
>> Why does S(5) = 40 rather than 13?
>
>... With S(5) = 13 we would have a non-prime cumulative sum for P(5) 
>    as the "1" of "13" summated to P(4) would give the composite 14. 
>
>Best (and sorry to be unclear),

But why are you allowing a(9) to equal a(3)=4? If I understand your
rules correctly, shouldn't you have

 S = 2,1,4,6,40,20,46,8,42,46,60,26,48,400,600, ???
 P = 2 3 7 1 11 11 22 3 44 45 55 66 77 888 888
           3 77 99 39 7 13 73 99 17 19 333 999

after which the sequence stops, as you can't get from 89 to 97 via
a single digit?

-- 
Chris Thompson
Email: cet1 at cam.ac.uk